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# Python | Check if list is sorted or not

• Difficulty Level : Easy
• Last Updated : 29 Dec, 2022

The sorted operation of list is essential operation in many application. But it takes best of O(nlogn) time complexity, hence one hopes to avoid this. So, to check if this is required or not, knowing if list is by default sorted or not, one can check if list is sorted or not. Lets discuss various ways this can be achieved.
Method #1 : Naive method The simplest way to check this is run a loop for first element and check if we could find any smaller element than it after that element, if yes, the list is not sorted.

## Python3

 `# Python3 code to demonstrate``# to check if list is sorted``# using naive method` `# initializing list``test_list ``=` `[``1``, ``4``, ``5``, ``8``, ``10``]` `# printing original list``print` `("Original ``list` `: " ``+` `str``(test_list))` `# using naive method to``# check sorted list``flag ``=` `0``i ``=` `1``while` `i < ``len``(test_list):``    ``if``(test_list[i] < test_list[i ``-` `1``]):``        ``flag ``=` `1``    ``i ``+``=` `1``    ` `# printing result``if` `(``not` `flag) :``    ``print` `("Yes, ``List` `is` `sorted``.")``else` `:``    ``print` `("No, ``List` `is` `not` `sorted``.")`

Output :

```Original list : [1, 4, 5, 8, 10]
Yes, List is sorted.```

Method #2 : Using sort() The new list can be made as a copy of the original list, sorting the new list and comparing with the old list will give us the result if sorting was required to get sorted list or not.

## Python3

 `# Python3 code to demonstrate``# to check if list is sorted``# using sort()` `# initializing list``test_list ``=` `[``10``, ``4``, ``5``, ``8``, ``10``]` `# printing original list``print` `("Original ``list` `: " ``+` `str``(test_list))` `# using sort() to``# check sorted list``flag ``=` `0``test_list1 ``=` `test_list[:]``test_list1.sort()``if` `(test_list1 ``=``=` `test_list):``    ``flag ``=` `1``    ` `# printing result``if` `(flag) :``    ``print` `("Yes, ``List` `is` `sorted``.")``else` `:``    ``print` `("No, ``List` `is` `not` `sorted``.")`

Output :

```Original list : [10, 4, 5, 8, 10]
No, List is not sorted.```

Method #3 : Using sorted() Using the similar analogy as the above method, but does not create a new space, but just a momentary space for that time and hence useful, shorter and faster method than above.

## Python3

 `# Python3 code to demonstrate``# to check if list is sorted``# using sorted()` `# initializing list``test_list ``=` `[``1``, ``4``, ``5``, ``8``, ``10``]` `# printing original list``print` `("Original ``list` `: " ``+` `str``(test_list))` `# using sorted() to``# check sorted list``flag ``=` `0``if``(test_list ``=``=` `sorted``(test_list)):``    ``flag ``=` `1``    ` `# printing result``if` `(flag) :``    ``print` `("Yes, ``List` `is` `sorted``.")``else` `:``    ``print` `("No, ``List` `is` `not` `sorted``.")`

Output :

```Original list : [1, 4, 5, 8, 10]
Yes, List is sorted.```

Method #4 : Using all() Most elegant, pythonic and faster way to check for sorted list is the use of all(). This uses the similar method as naive, but use of all() make it quicker.

## Python3

 `# Python3 code to demonstrate``# to check if list is sorted``# using all()` `# initializing list``test_list ``=` `[``9``, ``4``, ``5``, ``8``, ``10``]` `# printing original list``print` `("Original ``list` `: " ``+` `str``(test_list))` `# using all() to``# check sorted list``flag ``=` `0``if``(``all``(test_list[i] <``=` `test_list[i ``+` `1``] ``for` `i ``in` `range``(``len``(test_list)``-``1``))):``    ``flag ``=` `1``    ` `# printing result``if` `(flag) :``    ``print` `("Yes, ``List` `is` `sorted``.")``else` `:``    ``print` `("No, ``List` `is` `not` `sorted``.")`

Output :

```Original list : [9, 4, 5, 8, 10]
No, List is not sorted.```

Method #5 : Use the zip() function and the all() function

One approach that could be used to check if a list is sorted or not is to use the zip() function and the all() function. This approach involves pairing up the elements of the list in pairs using zip(), and then using all() to check if the elements of each pair are in the correct order.

Here is an example of how this approach could be implemented:

## Python3

 `# initialize the list``test_list ``=` `[``1``, ``4``, ``5``, ``8``, ``10``]` `# check if the list is sorted using zip() and all()``is_sorted ``=` `all``(a <``=` `b ``for` `a, b ``in` `zip``(test_list, test_list[``1``:]))` `# print the result``if` `is_sorted:``    ``print``(``"Yes, the list is sorted."``)``else``:``    ``print``(``"No, the list is not sorted."``)``#This code is contributed by Edula Vinay Kumar Reddy`

Output

`Yes, the list is sorted.`

The auxiliary space complexity of the zip() and all() approach to checking if a list is sorted is O(1), as it does not require any additional space to store data beyond the original list.

The time complexity of this approach is O(n), as it involves iterating through the elements of the list once to check if they are in the correct order.

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